Can someone help me with the following problem:
Consider the following hash function with separate chaining: It takes a non-negative integer n < 100000 and calculates its hash value as (2d1 + 3d2 + 5d3 + 7d4 + 11d5)%47, where % is modulo devision, d1 is the most significant digit, and d5 is the least significant one (the others are therefore in decreasing significance order). Suppose we have 2000 numbers. Prove that there exists a positive integer x for which we can find at least 43 numbers among the given 2000, whose hash value equals to x. Some hints to solve this problem: • Can you check how many different hash values can be? • Now, suppose that the statement is incorrect. Maximum how many numbers the hash table can store?
Easy Math Editor
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2^{34}
a_{i-1}
\frac{2}{3}
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\sum_{i=1}^3
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